## Calculus: Early Transcendentals 8th Edition

Use comparison test with $a_n=\frac{arctan n }{n^{2}}$ and $b_{n}=\frac{1}{n^{2}}$ $\lim\limits_{n \to \infty}\frac{a_{n}}{b_{n}}=\lim\limits_{n \to \infty}\frac{arctan n/n^{2} }{n^{2}}$ $=\lim\limits_{n \to \infty}arctan n$ $=\frac{\pi}{2}\ne 0 \ne \infty$ The given series absolutely converges because $\Sigma_{n=1 }^{\infty}\frac{1}{n^{2}}$ is a convergent p-series with $p=2\gt 1$ Hence, the series is absolutely convergent.